Question: $f(x) = \begin{cases} \dfrac{5}{x} & \text{for} ~~~~x\gt{5} \\ x-4& \text{for} ~~~~ x \leq5\end{cases}$ Evaluate the definite integral. $\int^6_{4}f(x)\,dx = $ Choose 1 answer: Choose 1 answer: (Choice A) A $ 5\ln\left(\dfrac54\right) +\dfrac32$ (Choice B) B $5\ln\left(\dfrac65\right) +\dfrac12$ (Choice C) C $ 6\ln\left(\dfrac56\right) -\dfrac12$ (Choice D) D Undefined
Explanation: Splitting up the definite integral Since we're working with a piecewise function, we need to split the definite integral up into two pieces: $\phantom{=} \int^6_{4}f(x)\,dx$ $= \int^6_{5}f(x)\,dx + \int^{5}_{4}f(x)\,dx~~~~~~$ [Why did we split at 5?] $= \int^6_{5}\dfrac5x\,dx + \int^{5}_{4}(x-4)\,dx ~~~~~~$ Evaluating each piece Next, let's evaluate each of these definite integrals one at a time. The first definite integral: $\begin{aligned} \int^6_{5}\dfrac5x\,dx &=5\ln(x)\Bigg|^6_{{5}} \\\\ &= \left[5\ln( 6) \right] - \left[5\ln ({5})\right] \\\\ &= {5\ln\left(\dfrac65\right)}\end{aligned}$ The second definite integral: $\begin{aligned} \int^{5}_{4}(x-4)\,dx &=\dfrac12x^2-4x\Bigg|^5_{{4}} \\\\ &= \left[\dfrac12\cdot\left({5} \right)^2 - 4\cdot(5) \right] - \left[ \dfrac12\cdot\left({4} \right)^2 - 4\cdot(4)\right] \\\\ &= \left[-\dfrac{15}2\right] -\left[-8 \right] \\\\ &= {\dfrac12}\end{aligned}$ Putting the pieces together Now let's add these two pieces together to find the answer: $\phantom{=} \int^6_{5}\dfrac5x\,dx + \int^{5}_{4}(x-4)\,dx$ $ = {5\ln\left(\dfrac65\right)} + {\dfrac12}$ The answer $\int^6_{4}f(x)\,dx = 5\ln\left(\dfrac65\right) +\dfrac12$